(x-4) (4x-3) +3 =0

To solve the equation \((x-4)(4x-3) + 3 = 0\), let's first simplify it step by step. 1. Expand the left side of the equation: \[ (x-4)(4x-3) = x(4x-3) - 4(4x-3) = 4x^2 - 3x - 16x + 12 = 4x^2 - 19x + 12 \] 2. Now rewrite the equation: \[ 4x^2 - 19x + 12 + 3 = 0 \] which simplifies to: \[ 4x^2 - 19x + 15 = 0 \] 3. Now we need to solve the quadratic equation \(4x^2 - 19x + 15 = 0\). We can use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -19\), and \(c = 15\). 4. Calculate the discriminant: \[ b^2 - 4ac = (-19)^2 - 4(4)(15) = 361 - 240 = 121 \] 5. Now apply the quadratic formula: \[ x = \frac{-(-19) \pm \sqrt{121}}{2(4)} = \frac{19 \pm 11}{8} \] 6. This gives us two potential solutions: \[ x = \frac{19 + 11}{8} = \frac{30}{8} = \frac{15}{4} \] and \[ x = \frac{19 - 11}{8} = \frac{8}{8} = 1 \] Thus, the solutions to the equation \((x-4)(4x-3) + 3 = 0\) are: \[ x = \frac{15}{4} \quad \text{and} \quad x = 1 \]